Lua plain searching with string.gsub?

4
votes

With Lua's string.find function, there is an optional fourth argument you can pass to enable plain searching. From the Lua wiki:

The pattern argument also allows more complex searches. See the PatternsTutorial for more information. We can turn off the pattern matching feature by using the optional fourth argument plain. plain takes a boolean value and must be preceeded by index. E.g.,

= string.find("Hello Lua user", "%su")         -- find a space character followed by "u"

10      11

= string.find("Hello Lua user", "%su", 1, true) -- turn on plain searches, now not found

nil

Basically, I was wondering how I can accomplish the same plain searching using Lua's string.gsub function.

3
stringlua
If you don't mind me asking, why are you trying to use gsub for something it wasn't at all meant for? - itdoesntwork
I'm trying to replace large chunks of a string with another string. - David
Oh, I made that comment assuming that there was a plaintext search-replace function already existing in the string library, but reading through the docs I'm not so sure it's there - itdoesntwork
I'm pretty sure there isn't :( - David
Escape every non-alphanumeric character in your search string. - Etan Reisner

3 Answers

2
votes

I expected there to be something in the standard library for this, but there isn't. The solution, then, is to escape the special characters in the pattern so they don't perform their usual functions.

Here's the general idea:

  1. obtain the pattern string
  2. replace any special characters with % followed by it (for example, % becomes %%, [ becomes %[
  3. use this as your search pattern for replacing the text
2
votes

Here is a simple library function for text replacement:

function string.replace(text, old, new)
  local b,e = text:find(old,1,true)
  if b==nil then
     return text
  else
     return text:sub(1,b-1) .. new .. text:sub(e+1)
  end
end

This function can be called as newtext = text:replace(old,new).

Note that this only replaces the first occurrence of old in text.

2
votes

Use this function to escape all magic characters (and only those) in your search string.

function escape_magic(s)
  return (s:gsub('[%^%$%(%)%%%.%[%]%*%+%-%?]','%%%1'))
end